Problem: Simplify the following expression: $y = \dfrac{-7x^2- 10x- 3}{-7x - 3}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(-7)}{(-3)} &=& 21 \\ {a} + {b} &=& &=& {-10} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $21$ and add them together. The factors that add up to ${-10}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-3}$ and ${b}$ is ${-7}$ $ \begin{eqnarray} {ab} &=& ({-3})({-7}) &=& 21 \\ {a} + {b} &=& {-3} + {-7} &=& -10 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({-7}x^2 {-3}x) + ({-7}x {-3}) $ Factor out the common factors: $ x(-7x - 3) + 1(-7x - 3)$ Now factor out $(-7x - 3)$ $ (-7x - 3)(x + 1)$ The original expression can therefore be written: $ \dfrac{(-7x - 3)(x + 1)}{-7x - 3}$ We are dividing by $-7x - 3$ , so $-7x - 3 \neq 0$ Therefore, $x \neq -\frac{3}{7}$ This leaves us with $x + 1; x \neq -\frac{3}{7}$.